Shouldn't these two Ruby snippets work the same way?

Question

In venturing into Ruby, I started toying with things like the way Ruby returns the last thing you've mentioned even if it was not after a return construct. However, why don't these two snippets work the same way? Shouldn't they?

module Enumerable
  def palindrome?
    reversed_self = self.reverse
    self.each_with_index {|el,index|
      unless self[index]==reversed_self[index]
        return false ## <-----
      end
    }
    true
  end
end

all good so far: puts ['foo','bar','baz'].palindrome? prints 'false'

and

module Enumerable
  def palindrome?
    reversed_self = self.reverse
    self.each_with_index {|el,index|
      unless self[index]==reversed_self[index]
        false ## <------
      end
    }
    true
  end
end

puts ['foo','bar','baz'].palindrome? prints 'true' for some reason

What's the science behind this?

Theo · Accepted Answer · 2012-03-11 13:35:52Z

3

Ruby will return the value of the last executed expression in a method. The false in the second version is not the last expression, there's nothing telling Ruby to stop executing at that point so it will chug along until the method ends.

return is a way to explicitly say to Ruby to stop executing and return a value.

answered Mar 11, 2012 at 13:35

Theo

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Community · Accepted Answer · 2017-05-23 12:11:25Z

2

Not quite! A return from inside a block is different from a return inside a lambda, as mentioned in my answer here. When you return from inside a block, you're returning from the entire method rather than just the block.

We can illustrate this as follows:

return :foo # => LocalJumpError: unexpected return
[1, 2, 3].map { return :foo } # => LocalJumpError: unexpected return
[1, 2, 3].map { :foo } # => [:foo, :foo, :foo]

Normally, this doesn't happen with lambdas:

l = lambda { return :foo }
l.call # => :foo
[1, 2, 3].map { l.call } # => [:foo, :foo, :foo]

But when we try to pass the lambda as a block to the method, the behavior changes back:

[1, 2, 3].map &l # => LocalJumpError: unexpected return

edited May 23, 2017 at 12:11

CommunityBot

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answered Mar 11, 2012 at 13:34

jtbandes

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2 Comments

Felipe Over a year ago

So you mean that when there's a return clause inside a block we return from the block only? I think you got confused in your answer?

jtbandes Over a year ago

No, it's exactly the opposite. A return from inside the block tries to return from the enclosing method. I edited my second sentence to clarify.

Sergio Tulentsev · Accepted Answer · 2012-03-11 13:33:06Z

1

If no return statement is present, then return value of a function is the last value evaluated. In the second snipped the last value is always true.

First snippet returns early with false. Second does nothing with that false, it's discarded.

answered Mar 11, 2012 at 13:33

Sergio Tulentsev

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4 Comments

jtbandes Over a year ago

It doesn't discard it: assuming the condition is satisfied, the unless..end block will evaluate to the false inside of it. The point is that the return statement will return from the whole method.

Sergio Tulentsev Over a year ago

@jtbandes: and then value of that unless is discarded :)

jtbandes Over a year ago

Well alright, it's discarded by the each, but not by a property of the language.

Sergio Tulentsev Over a year ago

@jtbandes: agree, I could have used better wording.

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